Answer :
The vapor pressure is 0.0023 torr.
The Clausius-Clapeyron equation, a formula frequently used to determine vapor pressure at a given temperature, must be employed to answer this question. Let's write the equation now that we know the mercury's enthalpy of vaporization:
Clausius Clapeyron equation:
Ln (P₂ / P₁) = (-ΔHv / R)(1/T₂ - 1/T₁) (1)
Where:
R: universal constant of gases (8.314 J / K.mol)
P₂: Vapour pressure at 23°C (or 296 K)
P₁: Pressure of mercury at the boiling point (1 atm)
T₂: the temperature at 23 °C
T₁: Boiling point of mercury (357 °C or 630 K)
We can reasonably infer that the pressure at this point is 1 atm since the boiling point of mercury has been provided. Keep in mind that when a substance boils, it does so because its internal pressure has reached the atmospheric pressure of 1 atm. Once this is established, all that is left to do is find P₂. We'll move very carefully through this process so you can comprehend it.
Now,
Ln (P₂ / 1) = (-59100 J/mol / 8.314 J / K.mol) (1/296 - 1/630)
Ln P₂ = -7108.49 * (3.378 x 10⁻³ - 1.59 x 10⁻³)
Ln P₂ = -7108.49 * (1.788 x 10⁻³)
Ln P₂ = -12.709
P₂ = 10^(-12.709)
P₂ = 3.02 x 10⁻⁶ atm
P₂ = 2.18x10⁻⁵ * 760 mm Hg
P₂ = 0.0023 mm Hg
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